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Proof in linear algebra

  1. Jun 2, 2009 #1
    1. The problem statement, all variables and given/known data
    Suppose S, T[tex]\in[/tex]the set of transformations from V to V is such
    that every subspace of V with dimension dimV-1 is invariant under T.
    Prove that T is a scalar multiple of the identity operator.


    2. Relevant equations

    T=[tex]\lambda[/tex]I

    3. The attempt at a solution
    u[tex]\in[/tex]U U[tex]\subset[/tex]V
    dimV=k dimU=k-1
    I[tex]\lambda[/tex]u=Tu
    Tu-I[tex]\lambda[/tex]u=0
    Since Iu=u
    and Tu=[tex]\lambda[/tex]u
    Tu=TIu=I[tex]\lambda[/tex]u
    So TIu=I[tex]\lambda[/tex]u
    with T=[tex]\lambda[/tex]
     
  2. jcsd
  3. Jun 2, 2009 #2

    Dick

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    You are assuming what you want to prove and then going in circles. Be concrete. Let dim(V)=n and pick a basis {v1,v2,...,vn}. Write e.g. T(v1)=a1*v1+a2*v2+...+an*vn. Can you show a2=0, a3=0, ... an=0? Consider the subspaces of V spanned by removing one of the v's from the basis. If you can then you can show T of any basis element is a multiple of itself. Now can you show the multiplication factor for any basis element is the same?
     
  4. Jun 2, 2009 #3
    Okay I'll use your basis {v1,v2,...,vn}
    Choose v in V.
    v=c1v1+....+cnvn
    I'll use T(v1)=T(c1v1)+....+T(cnvn)
    Consider a subspace U1 of V and v1 in U1
    For T to be a multiple of v1 we need
    T(v1)=[tex]\lambda[/tex]c1v1+0*c2v2+...+0*cnvn
    since we want to map from V to U1 and not U2 etc
    we need a multiple of v1 since Tv2 etc is not in U1
    So v2+...+vn is in nullT of v1
    T(v1)=[tex]\lambda[/tex]c1v1+0+...+0
    T(v1)=[tex]\lambda[/tex]c1v1
    Since Iv1=v1
    T(Iv1)=[tex]\lambda[/tex]Iv1
     
    Last edited: Jun 2, 2009
  5. Jun 2, 2009 #4

    Dick

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    That makes no sense whatsoever. Take the case n=3. T(v1)=c1*v1+c2*v2+c3*v3. v1 is in the span of {v1,v2} and that 2 dimensional subspace is invariant. So T(v1) is in the span of {v1,v2}. What does that tell you about c3? Why? Try not to just write a bunch of gibberish.
     
  6. Jun 2, 2009 #5
    c3 should be 0 because v3 is not in {v1, v2} but 0 is. So 0*v3=0
    and c1v1+c2v2+0=c1v1+c2v2 in the span {v1, v2}
    So to go from {v1, v2} back to itself
    with T(v1)=c1v1+c2v2+c3v3 with c3=0 since
    c3v3 is not in in the span of {v1, v2} if c3=/=0
     
  7. Jun 2, 2009 #6

    Mark44

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    Two specific examples of things that make no sense are these:
    T is a transformation, and [itex]\lambda[/itex] is an eigenvalue, a number. The two are incomparable.
    v1 is a vector. T operates on vectors, but isn't a vector, so can't be a scalar multiple of a vector.
     
  8. Jun 2, 2009 #7

    Dick

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    That's not super clear, but you have the right idea. T(v1) is in the span. The span of {v1,v2} is the set of all vectors a*v1+b*v2 and {v1,v2,v3} is a basis. So c3=0. Now think about the invariant subspace spanned by {v1,v3}.
     
    Last edited: Jun 2, 2009
  9. Jun 2, 2009 #8
    For {v1,v3}
    T(v3)=a1v1+a2v2+a3v3+a4v4
    with a2 and a4=0
    T(v3)=a1v1+0+a3v3+0=a1v1+a3v3 in {v1, v3}



    Lets give the proof another shot.
    We use the basis {v1....vn-1}
    T(v)=a1v1+...+an-1vn-1+anvn
    Setting this equation to 0
    0=a1v1+...+an-1vn-1+anvn
    -anvn=a1v1+...+an-1vn-1
    knowing that anvn[tex]\notin[/tex]{v1...vn-1}
    an=0
    so 0=a1v1+...+an-1vn-1
    since the elements of any basis are linear independent
    we can only have every "a" be equal to zero to get
    a1vn+....+an-1vn-1 to be=0
    So each T(vi)=aivi with ai=0
    Thus a1...an-1=0
    So T(v1)+...+T(vn-1)
    =0*(v1)+...+0*(vn-1)
    =0*(v1+....+vn-1)
     
    Last edited: Jun 2, 2009
  10. Jun 2, 2009 #9

    Dick

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    What is 'v'?? Why do you think you can set T(v)=0?? You are way off track.
     
  11. Jun 2, 2009 #10

    Dick

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    Stick with n=3. And stick with T(v1)=c1*v1+c2*v2+c3*v3. You've already shown c3=0 by considering the invariant subspace spanned by {v1,v2}. I hope. If not think about it again. Now consider the invariant subspace spanned by {v1,v3}.
     
  12. Jun 2, 2009 #11
    Oops, lol my mistake. The main space is only of dim3, v4 doesn't even exist here.
    Thats what you mean by n=3, right? One question. T(v1)=a1v1+a2v2+a3v3
    and we want to go from {1,3} to {1,3}.Though it is obvious that a2=0 since v2
    is not in{1,3}, we are only dealing with v1. What I mean is that
    T(v1) should be a1*v1 so shouldn't a3 be also 0? This will help me
    when I attempt the proof again (yeah I don't quit lol).
     
  13. Jun 2, 2009 #12

    Dick

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    Yes, that's what I mean by n=3. If you can really understand how the proof works for n=3 then you should be able to do it for any n. Sure the invariance of span{v1,v2} implies c3=0 and the invariance of span{v1,v3} implies c2=0. So T(v1)=c1*v1. There is nothing special about v1 vs v2 or v3. So clearly, T(v2)=c2*v2 and T(v3)=c3*v3. Don't attempt the proof again until you really get this. Otherwise you'll just post a sequence of random symbols, if past experience is any judge. Now you have to show c1=c2=c3. Big hint: span{v1+v2,v3} is an invariant subspace, since it has dimension 2.
     
  14. Jun 2, 2009 #13
    Alright just to be sure that I post something that makes sense
    :tongue2: am I seeing this clearly: we have the basis {v1, v3} a
    basis for some subspace for some space with basis {v1,v2,v3}
    We take T(v1)=c1v1+c2v2+c3v3. c2=0 which we know.
    But I would like to know if this is right: c3=0 because
    T(v1)=c1v1. T is only acting on v1. So we get T(v1)=T(v1)+T(v2)+T(v3)
    =c1v1+c2v2+c3v3=c1v1+0*v2+0*v3
    =c1v1+0+0=c1v1
     
  15. Jun 2, 2009 #14

    Dick

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    Ack! No, c3=0 BECAUSE T(v1) is in the span of {v1,v2} and span{v1,v2} is invariant since it's dimension=2. Not c3=0 BECAUSE T(v1)=c1*v1. We don't know that yet. You keep interchanging premise with consequence. Proofs mean you go from what you know to what you don't know. Not the reverse. No offence, but you seem to be kind of tone-deaf about this.
     
  16. Jun 2, 2009 #15
    Ok for the span {v1, v3}
    T(v1)=c1v1+c2v2+c3v3
    c2=0 since v2 is not in {v1, v3}
    which leaves T(v1)=c1v1+0+c3v3
    T(v1)=T(v1)+T(v3)
    T(v1)-T(v1)=T(v3)
    0=T(v3)
    0=c3v3
    since v3 is a member of the basis, it=/=0
    so c3=0
    I'm not sure about this one, so I won't continue it
    without reassurance.
     
  17. Jun 3, 2009 #16

    Dick

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    You are sort of ok this far. The language is pretty informal and you aren't giving all the justifications, but you gave a reason WHY c2=0. Good.

    Wrong. You changed c1*v1 into T(v1) and c3*v3 into T(v3) without giving any reason why. Because there is no reason why. You just changed it. Go back to before this misstep and tell me why c3=0. And don't write down ANYTHING without saying 'because' and giving a valid reason.
     
  18. Jun 3, 2009 #17
    Okay. Lets start from there.
    T(v1)=c1v1+0+c3v3
    Now mapping from {v1, v2} to {v1,v2}
    T(v1)=c1v1+c2v2+c3v3
    we found that c2v2=0 since c2=0
    and because c3v3 is not within the span {v1,v2}
    as it is not a linear combination of its vectors
    unless c3=0, c3 must be zero.
    Now, we found that c2 and c3 are both zero.
    So T(v1)=c1v1+0v2+0v3=c1v1
    Not done yet, just need to know if this is right.
     
  19. Jun 3, 2009 #18

    Dick

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    Ok, I'll buy that.
     
  20. Jun 3, 2009 #19
    Okay, we're moving somewhere.
    :smile:
    For convenience I will copy paste the second part of the proof and
    continue from there.

    T(v1)=c1v1+0+c3v3
    Now mapping from {v1, v2} to {v1,v2}
    T(v1)=c1v1+c2v2+c3v3
    we found that c2v2=0 since c2=0
    and because c3v3 is not within the span {v1,v2}
    as it is not a linear combination of its vectors
    unless c3=0, c3 must be zero.
    Now, we found that c2 and c3 are both zero.
    So T(v1)=c1v1+0v2+0v3=c1v1
    Now we take T(v2)=c1v1+c2v2+c3v3
    map from {v2, v3} to {v2, v3}
    c2v2+c3v3 is a linear combination in {v2,v3}
    Since we have found that T(v1)=c1v1
    and c1v1 is not within {v2,v3}, c1=0 for the
    same reasons as c2 and c3 are.
    Thus T(v2)=0*v1+c2v2+0*v1
    T(v2)=c2v2
    Since c2=0, T(v2)=0*v2=0
    T(v2)=0
    Now that we found that c1=0,
    T(v1)=c1*v1
    T(v1)=0*v1
    T(v1)=0
    Now mapping from {1,3} to {1,3}
    T(v3)=c1v1+c2v2+c3v3
    Since we have shown that c1 and c2 are 0,
    T(v3)=0*v1+0*v2+c3v3
    T(v3)=c3v3
    Since we have shown that c3=0
    T(v3)=0*v3
    T(v3)=0
    So T(v1)=c1v1+c2v2+c3v3
    T(v1)=T(v1)+T(v2)+T(v3)
    Since T(v1),T(v2),T(v3) all =0
    T(v1)=T(v1+v2+v3)
    T(v1)=0*(v1+v2+v3)
    can be shown for the others
     
  21. Jun 3, 2009 #20

    Dick

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    No, no, no. When you write T(v1)=c1*v1+c2*v2+c3*v3 and T(v2)=c1*v1+c2*v2+c3*v3 there is no implication that the c1 in T(v1) is the same as the c1 in T(v2). Otherwise, we'd be assuming T(v1)=T(v2). And we can't assume that. If it's causing too much confusion write T(v2)=d1*v1+d2*v2+d3*v3 and tell me what you can conclude about the d's.
     
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